Answer:
Hello there Dude answer is B :D hope it helped mark me brainliest.
Answer:
0.0002°, 0.1691°, 0.338°
Explanation:
Difference between the two line = 5.97 * 10-⁸m
d = 1 / N
N = 5.0 * 10³
d = 2.0 * 10⁴m
nL = Nsin¤
For first order
588.995 * 10-⁹ = 2.0 * 10-⁴ sin ¤
Sin¤ = 2.944*10-³
¤ = sin-¹ 0.002944
¤ = 0.1687°
First order ¤ =
Sin-¹(589.592*-⁹ / 2.0 * 10-⁴)
Sin-¹ (0.002947) = 0.1689°
Angular separation = 0.1689 - 0.1687 = 0.0002°
Second order ¤ = sin-¹ [2 (589.59*10-⁹ / 2.0*10-⁴)] = sin-¹ (0.005895)
Second order ¤ = 0.3378°
Angular difference = 0.3378° - 0.1687° = 0.1691°
Third order ¤ = sin-¹ [3(589.59*10-⁹ /2.0*10-⁴] = 0.5067°
Angular difference = 0.5067° - 0.1687° = 0.338°
Answer:
E = k Q 1 / (x₀-x₂) (x₀-x₁)
Explanation:
The electric field is given by
dE = k dq / r²
In this case as we have a continuous load distribution we can use the concept of linear density
λ= Q / x = dq / dx
dq = λ dx
We substitute in the equation
∫ dE = k ∫ λ dx / x²
We integrate
E = k λ (-1 / x)
We evaluate between the lower limits x = x₀- x₂ and higher x = x₀-x₁
E = k λ (-1 / x₀-x₁ + 1 / x₀-x₂)
E = k λ (x₂ -x₁) / (x₀-x₂) (x₀-x₁)
We replace the density
E = k (Q / (x₂-x₁)) [(x₂-x₁) / (x₀-x₂) (x₀-x₁)]
E = k Q 1 / (x₀-x₂) (x₀-x₁)
The change in electric potential energy of the ion is equal to the charge multiplied by the voltage difference:
where the charge q of the na+ ion is equal to one positive charge, so it's equal to the proton charge:
, and Vf and Vi are the final and initial voltages.
Substituting the numbers, we find:
Use Scoratic it works with any time of subject
